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enter image description here The points $(3,0)$, $(x,0)$, $(x,\frac{1}{x^2})$, and $(3,\frac{1}{x^2})$ are the vertices of a rectangle where $x\geq3$, as shown in the figure above. For what value of $x$ does the rectangle have a maximum area?

I know that $A'(x)=\frac{-1}{x^2}+\frac{6}{x^3}$ and when we set it equal to zero we get:

$\frac{-x+6}{x^3}=0$ $\implies x=6$

But then in the solution it also finds $A''(x)=\frac{2}{x^3}-\frac{18}{x^4}$ and says that since $A''(6)<0$, therefore the area is maximum at $x=6$

I'm just confused about the relevance of finding $A''(6)$. I know that the second derivative at a point reveals whether the function is concave up or down depending on if it is negative or positive, but I don't see how that fits into this question.

user130306
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    The solution provided is for completeness - from what you have done you have demonstrated that at $x=6$, the area function for this rectangle is either at a local maxima, local minima, or a neutral point. This does not say whether $A(6)$ is the maximum area - the second derivative test proves that A(6) is indeed a maximum – Dhanvi Sreenivasan Apr 20 '21 at 04:43

2 Answers2

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Note that $f’(x)=0$ does not imply the function has a local max at $x$ , for example consider $f(x)=x^3$ at $x=0$.
However the second derivative test states that if $f’(x)=0$ and $f’’(x)<0$ then $f$ has a local max at $x$ .
You still have to prove that this local max is actually a global max , this can be done by showing that the derivative is positive for all $x<6$ and negative for all $x>6$.

Vivaan Daga
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Equating the first derivative to 0 gives the point where the derivative is 0 ,i.e, points where there could br either Maxima or minima .

Lets say , the point comes out to be (a,b)

Now , of we have to determine that if the point is Maxima/minima , we check the double-derivative of the function at that point .

Therefore , if d²y/dx² is positive , then there's a MINIMA at that point.

And , if d²y/dx² is negative , it's a MAXIMA at that point.